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(1)    C is an eigenvalue  of T (the  set of these is called the point-
spectrum of T) ;

(2)    f is not an eigenvalue of T (which implies that T — £/ is a bijective
mapping of dom(T) onto a subspace L of E), the subspace L is dense in E, but
(jT — C/)"1 is not continuous on L (the set of these values is called the con-
tinuous spectrum of T) ;

(3)    C is not an eigenvalue of T and the subspace L is not dense in E (the
set of these values is called the residual spectrum of T).

For example, for the operator defined in (11.1.1), the point-spectrum is
empty, the continuous spectrum is the unit circle U : |C| = 1, and the residual
spectrum is the open disk |C| < 1 (Section 11.1, Problem 4).

(15.12.11)    The spectrum of a closed operator T on E is closed in C, and the
mapping C^^r(0 ofC - Sp(T) into £(E) is analytic.

Let Co e C — Sp(T), and put a = ||jRT(Co)ll- Then the continuous operator
= (/- (C - Co)JĞr(Co))"%(Co)isdefinedforallCsatisfying|C--C0|<fl-l>
and is analytic in this disk (83.2.1). For each x e dom(T), we have

*T(Co)(^ - C/) ' * = * + (Co -
and therefore V(Q(T - U) - x = x. For each integer n^Q, let

so that K(C) is the limit of the sequence (Fn(Q) in Jz?(E). By definition, we have
Fn(Q • y e dom(r) for all y e E and all n > 0, and

(T - CTOC) -y = (C - C0)K,(0 • y + (T - C

hence (r — CO^AO ' J tends to y as n -ğ +00. Since by hypothesis 7 is closed,
this implies that F(0 • y e dom(T) and (T - C/) P(0 • y = y- This shows that
T — C/ is a bijective linear mapping of dom(T) onto E and that the inverse
mapping is K(0. Hence C # Sp(T) and F(Q = ^r(0> which completes the

(15.12.12) £#£ N be a (not necessarily bounded) normal operator on E. With
the notation of (15. 12.8), we have Sp(N) = (J Sp(7V,,); the point-spectrum ofN


is the union of the point-spectra of the Nn , and the residual spectrum of N is
empty. For N to be self-adjoint, it is necessary and sufficient that its spectrum
should be real.d if the inverse linear